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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.3.21

In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ln(e^(θ)/(1+e^θ))

Guida verificata passo dopo passo
1
Rewrite the given function to simplify the expression inside the logarithm: \(y = \ln\left( \frac{e^{\theta}}{1 + e^{\theta}} \right)\).
Use the logarithm property \(\ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b)\) to separate the function into \(y = \ln(e^{\theta}) - \ln(1 + e^{\theta})\).
Simplify \(\ln(e^{\theta})\) using the property \(\ln(e^{x}) = x\), so the function becomes \(y = \theta - \ln(1 + e^{\theta})\).
Differentiate each term with respect to \(\theta\): the derivative of \(\theta\) is 1, and for \(-\ln(1 + e^{\theta})\), apply the chain rule.
For the second term, use the chain rule: \(\frac{d}{d\theta} \left[ -\ln(1 + e^{\theta}) \right] = - \frac{1}{1 + e^{\theta}} \cdot \frac{d}{d\theta} (1 + e^{\theta}) = - \frac{e^{\theta}}{1 + e^{\theta}}\).

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